Optimal. Leaf size=256 \[ -\frac {2 (B d-A e) (a+b x)^{7/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac {2 (7 b B d-2 A b e-5 a B e) (a+b x)^{5/2}}{15 e^2 (b d-a e) (d+e x)^{3/2}}-\frac {2 b (7 b B d-2 A b e-5 a B e) (a+b x)^{3/2}}{3 e^3 (b d-a e) \sqrt {d+e x}}+\frac {b^2 (7 b B d-2 A b e-5 a B e) \sqrt {a+b x} \sqrt {d+e x}}{e^4 (b d-a e)}-\frac {b^{3/2} (7 b B d-2 A b e-5 a B e) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{9/2}} \]
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Rubi [A]
time = 0.12, antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {79, 49, 52, 65,
223, 212} \begin {gather*} -\frac {b^{3/2} (-5 a B e-2 A b e+7 b B d) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{9/2}}+\frac {b^2 \sqrt {a+b x} \sqrt {d+e x} (-5 a B e-2 A b e+7 b B d)}{e^4 (b d-a e)}-\frac {2 b (a+b x)^{3/2} (-5 a B e-2 A b e+7 b B d)}{3 e^3 \sqrt {d+e x} (b d-a e)}-\frac {2 (a+b x)^{5/2} (-5 a B e-2 A b e+7 b B d)}{15 e^2 (d+e x)^{3/2} (b d-a e)}-\frac {2 (a+b x)^{7/2} (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)} \end {gather*}
Antiderivative was successfully verified.
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Rule 49
Rule 52
Rule 65
Rule 79
Rule 212
Rule 223
Rubi steps
\begin {align*} \int \frac {(a+b x)^{5/2} (A+B x)}{(d+e x)^{7/2}} \, dx &=-\frac {2 (B d-A e) (a+b x)^{7/2}}{5 e (b d-a e) (d+e x)^{5/2}}+\frac {(7 b B d-2 A b e-5 a B e) \int \frac {(a+b x)^{5/2}}{(d+e x)^{5/2}} \, dx}{5 e (b d-a e)}\\ &=-\frac {2 (B d-A e) (a+b x)^{7/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac {2 (7 b B d-2 A b e-5 a B e) (a+b x)^{5/2}}{15 e^2 (b d-a e) (d+e x)^{3/2}}+\frac {(b (7 b B d-2 A b e-5 a B e)) \int \frac {(a+b x)^{3/2}}{(d+e x)^{3/2}} \, dx}{3 e^2 (b d-a e)}\\ &=-\frac {2 (B d-A e) (a+b x)^{7/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac {2 (7 b B d-2 A b e-5 a B e) (a+b x)^{5/2}}{15 e^2 (b d-a e) (d+e x)^{3/2}}-\frac {2 b (7 b B d-2 A b e-5 a B e) (a+b x)^{3/2}}{3 e^3 (b d-a e) \sqrt {d+e x}}+\frac {\left (b^2 (7 b B d-2 A b e-5 a B e)\right ) \int \frac {\sqrt {a+b x}}{\sqrt {d+e x}} \, dx}{e^3 (b d-a e)}\\ &=-\frac {2 (B d-A e) (a+b x)^{7/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac {2 (7 b B d-2 A b e-5 a B e) (a+b x)^{5/2}}{15 e^2 (b d-a e) (d+e x)^{3/2}}-\frac {2 b (7 b B d-2 A b e-5 a B e) (a+b x)^{3/2}}{3 e^3 (b d-a e) \sqrt {d+e x}}+\frac {b^2 (7 b B d-2 A b e-5 a B e) \sqrt {a+b x} \sqrt {d+e x}}{e^4 (b d-a e)}-\frac {\left (b^2 (7 b B d-2 A b e-5 a B e)\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {d+e x}} \, dx}{2 e^4}\\ &=-\frac {2 (B d-A e) (a+b x)^{7/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac {2 (7 b B d-2 A b e-5 a B e) (a+b x)^{5/2}}{15 e^2 (b d-a e) (d+e x)^{3/2}}-\frac {2 b (7 b B d-2 A b e-5 a B e) (a+b x)^{3/2}}{3 e^3 (b d-a e) \sqrt {d+e x}}+\frac {b^2 (7 b B d-2 A b e-5 a B e) \sqrt {a+b x} \sqrt {d+e x}}{e^4 (b d-a e)}-\frac {(b (7 b B d-2 A b e-5 a B e)) \text {Subst}\left (\int \frac {1}{\sqrt {d-\frac {a e}{b}+\frac {e x^2}{b}}} \, dx,x,\sqrt {a+b x}\right )}{e^4}\\ &=-\frac {2 (B d-A e) (a+b x)^{7/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac {2 (7 b B d-2 A b e-5 a B e) (a+b x)^{5/2}}{15 e^2 (b d-a e) (d+e x)^{3/2}}-\frac {2 b (7 b B d-2 A b e-5 a B e) (a+b x)^{3/2}}{3 e^3 (b d-a e) \sqrt {d+e x}}+\frac {b^2 (7 b B d-2 A b e-5 a B e) \sqrt {a+b x} \sqrt {d+e x}}{e^4 (b d-a e)}-\frac {(b (7 b B d-2 A b e-5 a B e)) \text {Subst}\left (\int \frac {1}{1-\frac {e x^2}{b}} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {d+e x}}\right )}{e^4}\\ &=-\frac {2 (B d-A e) (a+b x)^{7/2}}{5 e (b d-a e) (d+e x)^{5/2}}-\frac {2 (7 b B d-2 A b e-5 a B e) (a+b x)^{5/2}}{15 e^2 (b d-a e) (d+e x)^{3/2}}-\frac {2 b (7 b B d-2 A b e-5 a B e) (a+b x)^{3/2}}{3 e^3 (b d-a e) \sqrt {d+e x}}+\frac {b^2 (7 b B d-2 A b e-5 a B e) \sqrt {a+b x} \sqrt {d+e x}}{e^4 (b d-a e)}-\frac {b^{3/2} (7 b B d-2 A b e-5 a B e) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{e^{9/2}}\\ \end {align*}
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Mathematica [A]
time = 0.42, size = 205, normalized size = 0.80 \begin {gather*} -\frac {\sqrt {a+b x} \left (2 a^2 e^2 (2 B d+3 A e+5 B e x)+2 a b e \left (A e (5 d+11 e x)+B \left (20 d^2+49 d e x+35 e^2 x^2\right )\right )+b^2 \left (2 A e \left (15 d^2+35 d e x+23 e^2 x^2\right )-B \left (105 d^3+245 d^2 e x+161 d e^2 x^2+15 e^3 x^3\right )\right )\right )}{15 e^4 (d+e x)^{5/2}}+\frac {b^{3/2} (-7 b B d+2 A b e+5 a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {e} \sqrt {a+b x}}\right )}{e^{9/2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1091\) vs.
\(2(222)=444\).
time = 0.09, size = 1092, normalized size = 4.27
method | result | size |
default | \(\frac {\left (90 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{3} d^{2} e^{2} x +30 B \,b^{2} e^{3} x^{3} \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}-60 A \,b^{2} d^{2} e \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}+90 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{3} d \,e^{3} x^{2}+75 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a \,b^{2} e^{4} x^{3}-105 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{3} d \,e^{3} x^{3}-315 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{3} d^{2} e^{2} x^{2}-92 A \,b^{2} e^{3} x^{2} \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}-315 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{3} d^{3} e x +75 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a \,b^{2} d^{3} e -20 B \,a^{2} e^{3} x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}-8 B \,a^{2} d \,e^{2} \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}-105 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{3} d^{4}-140 A \,b^{2} d \,e^{2} x \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}-20 A a b d \,e^{2} \sqrt {b e}\, \sqrt {\left (b x +a \right ) \left (e x +d \right )}-12 A \,a^{2} e^{3} \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+210 B \,b^{2} d^{3} \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}-196 B a b d \,e^{2} x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+30 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{3} d^{3} e +225 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a \,b^{2} d \,e^{3} x^{2}+225 B \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) a \,b^{2} d^{2} e^{2} x -140 B a b \,e^{3} x^{2} \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+322 B \,b^{2} d \,e^{2} x^{2} \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}-44 A a b \,e^{3} x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+490 B \,b^{2} d^{2} e x \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}-80 B a b \,d^{2} e \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+30 A \ln \left (\frac {2 b e x +2 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}+a e +b d}{2 \sqrt {b e}}\right ) b^{3} e^{4} x^{3}\right ) \sqrt {b x +a}}{30 \sqrt {\left (b x +a \right ) \left (e x +d \right )}\, \sqrt {b e}\, \left (e x +d \right )^{\frac {5}{2}} e^{4}}\) | \(1092\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 4.87, size = 798, normalized size = 3.12 \begin {gather*} \left [-\frac {15 \, {\left (7 \, B b^{2} d^{4} - {\left (5 \, B a b + 2 \, A b^{2}\right )} x^{3} e^{4} + {\left (7 \, B b^{2} d x^{3} - 3 \, {\left (5 \, B a b + 2 \, A b^{2}\right )} d x^{2}\right )} e^{3} + 3 \, {\left (7 \, B b^{2} d^{2} x^{2} - {\left (5 \, B a b + 2 \, A b^{2}\right )} d^{2} x\right )} e^{2} + {\left (21 \, B b^{2} d^{3} x - {\left (5 \, B a b + 2 \, A b^{2}\right )} d^{3}\right )} e\right )} \sqrt {b} e^{\left (-\frac {1}{2}\right )} \log \left (b^{2} d^{2} + 4 \, {\left (b d e + {\left (2 \, b x + a\right )} e^{2}\right )} \sqrt {b x + a} \sqrt {x e + d} \sqrt {b} e^{\left (-\frac {1}{2}\right )} + {\left (8 \, b^{2} x^{2} + 8 \, a b x + a^{2}\right )} e^{2} + 2 \, {\left (4 \, b^{2} d x + 3 \, a b d\right )} e\right ) - 4 \, {\left (105 \, B b^{2} d^{3} + {\left (15 \, B b^{2} x^{3} - 6 \, A a^{2} - 2 \, {\left (35 \, B a b + 23 \, A b^{2}\right )} x^{2} - 2 \, {\left (5 \, B a^{2} + 11 \, A a b\right )} x\right )} e^{3} + {\left (161 \, B b^{2} d x^{2} - 14 \, {\left (7 \, B a b + 5 \, A b^{2}\right )} d x - 2 \, {\left (2 \, B a^{2} + 5 \, A a b\right )} d\right )} e^{2} + 5 \, {\left (49 \, B b^{2} d^{2} x - 2 \, {\left (4 \, B a b + 3 \, A b^{2}\right )} d^{2}\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d}}{60 \, {\left (x^{3} e^{7} + 3 \, d x^{2} e^{6} + 3 \, d^{2} x e^{5} + d^{3} e^{4}\right )}}, \frac {15 \, {\left (7 \, B b^{2} d^{4} - {\left (5 \, B a b + 2 \, A b^{2}\right )} x^{3} e^{4} + {\left (7 \, B b^{2} d x^{3} - 3 \, {\left (5 \, B a b + 2 \, A b^{2}\right )} d x^{2}\right )} e^{3} + 3 \, {\left (7 \, B b^{2} d^{2} x^{2} - {\left (5 \, B a b + 2 \, A b^{2}\right )} d^{2} x\right )} e^{2} + {\left (21 \, B b^{2} d^{3} x - {\left (5 \, B a b + 2 \, A b^{2}\right )} d^{3}\right )} e\right )} \sqrt {-b e^{\left (-1\right )}} \arctan \left (\frac {{\left (b d + {\left (2 \, b x + a\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d} \sqrt {-b e^{\left (-1\right )}}}{2 \, {\left (b^{2} d x + a b d + {\left (b^{2} x^{2} + a b x\right )} e\right )}}\right ) + 2 \, {\left (105 \, B b^{2} d^{3} + {\left (15 \, B b^{2} x^{3} - 6 \, A a^{2} - 2 \, {\left (35 \, B a b + 23 \, A b^{2}\right )} x^{2} - 2 \, {\left (5 \, B a^{2} + 11 \, A a b\right )} x\right )} e^{3} + {\left (161 \, B b^{2} d x^{2} - 14 \, {\left (7 \, B a b + 5 \, A b^{2}\right )} d x - 2 \, {\left (2 \, B a^{2} + 5 \, A a b\right )} d\right )} e^{2} + 5 \, {\left (49 \, B b^{2} d^{2} x - 2 \, {\left (4 \, B a b + 3 \, A b^{2}\right )} d^{2}\right )} e\right )} \sqrt {b x + a} \sqrt {x e + d}}{30 \, {\left (x^{3} e^{7} + 3 \, d x^{2} e^{6} + 3 \, d^{2} x e^{5} + d^{3} e^{4}\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 668 vs.
\(2 (234) = 468\).
time = 1.93, size = 668, normalized size = 2.61 \begin {gather*} \frac {{\left (7 \, B b^{2} d {\left | b \right |} - 5 \, B a b {\left | b \right |} e - 2 \, A b^{2} {\left | b \right |} e\right )} e^{\left (-\frac {9}{2}\right )} \log \left ({\left | -\sqrt {b x + a} \sqrt {b} e^{\frac {1}{2}} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b}} + \frac {{\left ({\left ({\left (b x + a\right )} {\left (\frac {15 \, {\left (B b^{9} d^{2} {\left | b \right |} e^{6} - 2 \, B a b^{8} d {\left | b \right |} e^{7} + B a^{2} b^{7} {\left | b \right |} e^{8}\right )} {\left (b x + a\right )}}{b^{6} d^{2} e^{7} - 2 \, a b^{5} d e^{8} + a^{2} b^{4} e^{9}} + \frac {23 \, {\left (7 \, B b^{10} d^{3} {\left | b \right |} e^{5} - 19 \, B a b^{9} d^{2} {\left | b \right |} e^{6} - 2 \, A b^{10} d^{2} {\left | b \right |} e^{6} + 17 \, B a^{2} b^{8} d {\left | b \right |} e^{7} + 4 \, A a b^{9} d {\left | b \right |} e^{7} - 5 \, B a^{3} b^{7} {\left | b \right |} e^{8} - 2 \, A a^{2} b^{8} {\left | b \right |} e^{8}\right )}}{b^{6} d^{2} e^{7} - 2 \, a b^{5} d e^{8} + a^{2} b^{4} e^{9}}\right )} + \frac {35 \, {\left (7 \, B b^{11} d^{4} {\left | b \right |} e^{4} - 26 \, B a b^{10} d^{3} {\left | b \right |} e^{5} - 2 \, A b^{11} d^{3} {\left | b \right |} e^{5} + 36 \, B a^{2} b^{9} d^{2} {\left | b \right |} e^{6} + 6 \, A a b^{10} d^{2} {\left | b \right |} e^{6} - 22 \, B a^{3} b^{8} d {\left | b \right |} e^{7} - 6 \, A a^{2} b^{9} d {\left | b \right |} e^{7} + 5 \, B a^{4} b^{7} {\left | b \right |} e^{8} + 2 \, A a^{3} b^{8} {\left | b \right |} e^{8}\right )}}{b^{6} d^{2} e^{7} - 2 \, a b^{5} d e^{8} + a^{2} b^{4} e^{9}}\right )} {\left (b x + a\right )} + \frac {15 \, {\left (7 \, B b^{12} d^{5} {\left | b \right |} e^{3} - 33 \, B a b^{11} d^{4} {\left | b \right |} e^{4} - 2 \, A b^{12} d^{4} {\left | b \right |} e^{4} + 62 \, B a^{2} b^{10} d^{3} {\left | b \right |} e^{5} + 8 \, A a b^{11} d^{3} {\left | b \right |} e^{5} - 58 \, B a^{3} b^{9} d^{2} {\left | b \right |} e^{6} - 12 \, A a^{2} b^{10} d^{2} {\left | b \right |} e^{6} + 27 \, B a^{4} b^{8} d {\left | b \right |} e^{7} + 8 \, A a^{3} b^{9} d {\left | b \right |} e^{7} - 5 \, B a^{5} b^{7} {\left | b \right |} e^{8} - 2 \, A a^{4} b^{8} {\left | b \right |} e^{8}\right )}}{b^{6} d^{2} e^{7} - 2 \, a b^{5} d e^{8} + a^{2} b^{4} e^{9}}\right )} \sqrt {b x + a}}{15 \, {\left (b^{2} d + {\left (b x + a\right )} b e - a b e\right )}^{\frac {5}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{5/2}}{{\left (d+e\,x\right )}^{7/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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